Is Cramer's rule efficient for computational point of view?

I am not sure if Cramer's rule is used for computation purposes. Your help would mean a lot. Thanks!

• no. check some linear analisys book. like burden or Kincaid – Luis Felipe Oct 6 '15 at 11:56
• It has rather theoretical importance. It is used to prove some theorems in linear algebra. – user137035 Oct 6 '15 at 12:00
• See this as well. – J. M. is a poor mathematician Oct 6 '15 at 14:43
• Cramer's rule is often handy for small, fixed-size matrices like 3x3. I use it in computer graphics work occasionally. – Tavian Barnes Oct 6 '15 at 17:41
• Actually I would go a step further than "don't use Cramer's rule". If you are calculating the determinant or the inverse of a matrix as part of a bigger algorithm, that is almost always a bad thing to do. Mathematical formulas using matrix inverses often look neat, but that doesn't mean they are the best way to calculate something. – alephzero Oct 6 '15 at 18:16

Cramer is highly inefficient, of time complexity $O(n! \times n)$ with a naive determinant-finding algorithm, and $O(n^4)$ with e.g. LU decomposition. Gaussian elimination has cubic complexity.
• That's only if the deteminant is computer by definition. It may be computed by LU decomposition as well, in which case the complexity would be $O(n^4)$. – Sasha Oct 6 '15 at 11:56
• On a tangent: is there a difference between $O(n! \times n)$ and $O(n!)$? I'm thinking $O(n! \times n) \approx O((n+1)!) \approx O(n!)$? – Nick T Oct 6 '15 at 21:06
• $O((n+1)!)$ is not $O(n!)$ any more than $O(n^{k+1})$ is $O(n^k)$. – R.. Oct 6 '15 at 21:22